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Physics · Ch 2 — Mathematical Methods

Vector Product (cross product)

2.5.2

Vector Product (cross product)

The vector product (or cross product) of two vectors P⃗\vec P and Q⃗\vec Q is a VECTOR whose magnitude equals the product of their magnitudes and the sine of the smaller angle θ\theta between them, and whose direction is perpendicular to the plane containing P⃗\vec P and Q⃗\vec Q, given by the right-hand screw rule:

R⃗=P⃗×Q⃗=PQsin⁡θ u^r— (2.18)\vec R = \vec P\times\vec Q = PQ\sin\theta\,\hat u_r \qquad \text{--- (2.18)}

By the right-hand screw rule, if a screw is turned from P⃗\vec P towards Q⃗\vec Q through the smaller angle between them, the direction the screw tip advances is the direction of R⃗\vec R.

Properties.

  1. NOT commutative — anti-commutative instead: P⃗×Q⃗≠Q⃗×P⃗\vec P\times\vec Q \ne \vec Q\times\vec P; in fact Q⃗×P⃗=−(P⃗×Q⃗)\vec Q\times\vec P = -(\vec P\times\vec Q) --- (2.19), (2.20). The two products have the same magnitude but opposite directions.
  2. Distributive: A⃗×(B⃗+C⃗)=A⃗×B⃗+A⃗×C⃗\vec A\times(\vec B+\vec C) = \vec A\times\vec B+\vec A\times\vec C --- (2.21).
  3. Special angles: ∣P⃗×Q⃗∣=PQsin⁡θ|\vec P\times\vec Q| = PQ\sin\theta --- (2.22). If θ=0\theta=0 (parallel vectors) or θ=180∘\theta=180^\circ (anti-parallel), sin⁡θ=0\sin\theta=0 so the cross product is the zero vector. If θ=90∘\theta=90^\circ (perpendicular vectors), ∣P⃗×Q⃗∣=PQ|\vec P\times\vec Q|=PQ (maximum). In particular, i^×j^=k^\hat i\times\hat j=\hat k, j^×k^=i^\hat j\times\hat k=\hat i, k^×i^=j^\hat k\times\hat i=\hat j; and i^×i^=j^×j^=k^×k^=0⃗\hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=\vec 0 (property 4, since P⃗=Q⃗\vec P=\vec Q makes θ=0\theta=0).

Component (determinant) formula. For P⃗=Pxi^+Pyj^+Pzk^\vec P=P_x\hat i+P_y\hat j+P_z\hat k and Q⃗=Qxi^+Qyj^+Qzk^\vec Q=Q_x\hat i+Q_y\hat j+Q_z\hat k,

P⃗×Q⃗=∣i^j^k^PxPyPzQxQyQz∣=(PyQz−PzQy)i^+(PzQx−PxQz)j^+(PxQy−PyQx)k^— (2.23)\vec P\times\vec Q = \begin{vmatrix}\hat i & \hat j & \hat k \\ P_x & P_y & P_z \\ Q_x & Q_y & Q_z\end{vmatrix} = (P_yQ_z-P_zQ_y)\hat i+(P_zQ_x-P_xQ_z)\hat j+(P_xQ_y-P_yQ_x)\hat k \qquad \text{--- (2.23)}

This follows by expanding P⃗×Q⃗\vec P\times\vec Q term by term using i^×i^=j^×j^=k^×k^=0\hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=0, i^×j^=−j^×i^=k^\hat i\times\hat j=-\hat j\times\hat i=\hat k, j^×k^=−k^×j^=i^\hat j\times\hat k=-\hat k\times\hat j=\hat i, k^×i^=−i^×k^=j^\hat k\times\hat i=-\hat i\times\hat k=\hat j.

Area interpretation. The magnitude of the cross product of two vectors equals the area of the parallelogram whose adjacent sides represent the two vectors — for a parallelogram with base ∣P⃗∣|\vec P| and the other side ∣Q⃗∣|\vec Q| inclined at angle θ\theta, the perpendicular height is ∣Q⃗∣sin⁡θ|\vec Q|\sin\theta, so the area (base × height) =PQsin⁡θ=∣P⃗×Q⃗∣= PQ\sin\theta = |\vec P\times\vec Q| --- (2.24). …

Figure 2.10Direction of the vector (cross) product by the right-hand screw rule

What this figure shows. Two-panel figure showing the right-hand-screw-rule direction of the cross product. Panel (a): vectors P⃗\vec P and Q⃗\vec Q are drawn from a common point O at angle θ\theta to each other, both lying in a horizontal plane; a third vector R⃗\vec R, labelled with a small circled-dot symbol at its tip, is drawn perpendicular to the plane of P and Q (pointing 'out of the page' toward the viewer), representing R⃗=P⃗×Q⃗\vec R=\vec P\times\vec Q found via the right-hand screw rule (screw turned from P towards Q through the smaller angle θ). Panel (b) shows the SAME two vectors P⃗\vec P and Q⃗\vec Q at the same angle θ, but with the resulting vector S⃗\vec S (labelled with a small crossed symbol at its tip, indicating it points 'into the page') representing S⃗=Q⃗×P⃗\vec S=\vec Q\times\vec P — pointing in the OPPOSITE direction to R⃗\vec R in panel (a), since reversing the or …

Figure 2.11Cross product magnitude as the area of a parallelogram

What this figure shows. A parallelogram is drawn with vertex O; one side OA⃗\vec{OA} represents vector P⃗\vec P (taken as the base of the parallelogram) and the adjacent side OB⃗\vec{OB} represents vector Q⃗\vec Q, inclined to P⃗\vec P at angle θ\theta. A perpendicular is dropped from B onto the line OA (or its extension), meeting it at point D; the segment BD, of length h, is marked as the perpendicular height of the parallelogram measured from base OA. This construction shows h=OBsin⁡θ=Qsin⁡θh = OB\sin\theta = Q\sin\theta, so the parallelogram's area (base × height =OA×BD=PQsin⁡θ= OA\times BD = PQ\sin\theta) is numerically equal to the magnitude of the cross product ∣P⃗×Q⃗∣|\vec P\times\vec Q|. Caption re …

Misc Ex.2.7Angular momentum as a cross product

Worked out. Given the position vector r⃗=4i^+6j^−3k^\vec r = 4\hat i+6\hat j-3\hat k and linear momentum vector p⃗=2i^+4j^−5k^\vec p = 2\hat i+4\hat j-5\hat k of a moving body, the problem asks for the angular momentum L⃗=r⃗×p⃗\vec L = \vec r\times\vec p about the origin. The method sets up the cross product as a 3×3 determinant with i^,j^,k^\hat i,\hat j,\hat k in the top row and the components of r⃗\vec r and p⃗\vec p in the next two rows, then expands it component-by-component — a direct application of the vector-product formula to the physically important quantity of angular momentum i …

Misc Ex.2.9Finding an unknown component for two same-direction vectors

Worked out. Given P⃗=4i^−j^+8k^\vec P = 4\hat i-\hat j+8\hat k and Q⃗=2i^−mj^+4k^\vec Q = 2\hat i-m\hat j+4\hat k, where mm is an unknown constant, the problem asks you to find the value of mm for which P⃗\vec P and Q⃗\vec Q point in the SAME direction. The method uses the fact that same-direction (parallel, θ=0) vectors have corresponding components in the same fixed ratio, i.e. Px/Qx=Py/Qy=Pz/QzP_x/Q_x = P_y/Q_y = P_z/Q_z; setting up this proportion using the known x- and z-components and solving for the ratio, then applying it to the y-components, g …

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